templates - C++ functor binding -

i tried use old bind2nd function in way:

template<typename t> class printer { public:   void operator()(t a, string& kd)   {         cout<<a<<endl;   } };   int main(int argc, char *argv[]) {    string nme = "j-dar";    auto f1 = bind2nd(printer<int>(),nme);     //f1(5);    return 0; } 

but lot of errors:

required here error: no type named 'first_argument_type' in 'class printer<int>'  class binder2nd        ^ error: no type named 'second_argument_type' in 'class printer<int>'    typename _operation::second_argument_type value;                                              ^ error: no type named 'second_argument_type' in 'class printer<int>'    binder2nd(const _operation& __x,    ^ error: no type named 'result_type' in 'class printer<int>'    operator()(const typename _operation::first_argument_type& __x) const    ^ error: no type named 'result_type' in 'class printer<int>'    operator()(typename      _operation::first_argument_type& __x) const    ^ required here error: no type named 'second_argument_type' in 'class printer<int>'    typedef typename _operation::second_argument_type _arg2_type;                                            

from can see it's correct don't know going on. ^

first of all: recommend using abandoning bind1st() , bind2nd(), deprecated in c+11, , in general obsolete support functional programming of c++03 standard library.

you should rather use c++11's std::bind(), since seems can afford - judging fact using auto keyword:

#include <functional>  // ...  auto f1 = std::bind(printer<int>(), std::placeholders::_1, nme); 

---

this said, record, deprecated std::bind2nd() function requires metadata signature of functor's call operator, , expects these metadata provided type aliases in functor class. instance:

template<typename t> class printer { public:      // these metadata must present in order bind1st , bind2nd work...     typedef void result_type;     typedef t first_argument_type;     typedef string const& second_argument_type;      void operator()(t a, string const& kd) const //                                         ^^^^^ // bonus advice #1: //                                               // , should //                                               // const-qualified //                              ^^^^^ //                              bonus advice #2: why not taking //                              reference const here? ;)     {         cout<<a<<endl;     } }; 

a simpler way of achieving above use (also deprecated) class template std::binary_function base class, , let class template define appropriate type aliases:

template<typename t> class printer : public std::binary_function<t, string const&, void> //              ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ { public:     void operator()(t a, string const& kd) const     {         cout<<a<<endl;     } }; 

but again, please consider putting std::bind1st(), std::bind2nd(), std::unary_function , std::binary_function, in drawer. superseded c++11's more powerful support functional programming.